Voltage divider - Looking for a reference with respect to Vout resistnce

[Femme_physics]

In voltage divider, how high must the resistance on Vout be so that voltage divider can be used according to convention? I'm looking for official references..

If anyone can help I'd appreciate it.

 

[sophiecentaur]

This is a 'practical' question? You would need to do the sums for each case.
The value of the 'load' resistor needs to be added in parallel with the resistor across which you are measuring the Vout. The 'bottom resistor' of the pair in the divider has been modified. The answer to "how high" that resistor must be depends totally upon the accuracy that you require.

If the load is 100 times the resistance of the potential divider resistor then the error will be about 1% - because the effective resistance has been reduced by about 1%.

 

[Femme_physics]

Do you have a resource for that?

 

[Kholdstare]

Why'd you need a resource for that? Is not the thing simple enough?

The relative error will be as follows:

\DeltaV/V = \frac{|V_{out-load} - V_{out}|}{V_{out}} = |\frac{R_{L}}{R_{L}+R_{1}||R_{2}}-1|

I guess now you can figure out that to reduce the error all you need to do is either make RL very large relative to R1||R2 or make R1||R2 very small relative to RL.

 

[Femme_physics]

If it's an open circuit-- how do u know there's no load? Open means that you don't know what's attached to it.

And if there's no load, doesn't it mean that the current won't bother to flow into R2 resistance? It'd much prefer to flow through the region where there is no load.

As far as to why I need the resource...I just want to do proper research of the information I get on the forum :)

 

[sophiecentaur]

Do you have a resource for that?

Any textbook is sufficient resource. What you need is actual experience of doing some calculations to see what happens.
I assume that you can work out 1. the equivalent resistor for two in parallel and I assume that the 2. potential divider equation is familiar.
Just substitute some numbers in there and you will get a feel for it. There is no definite answer as to 'how much' error is acceptable.

It'd much prefer to flow through the region where there is no load.

? Since when did current have any preferences? The expression 'No Load' means an Infinite Resistance so the voltage on the Potential Divider would not be affected.

 

[Femme_physics]

Any textbook is sufficient resource. What you need is actual experience of doing some calculations to see what happens.
I assume that you can work out 1. the equivalent resistor for two in parallel and I assume that the 2. potential divider equation is familiar.
Just substitute some numbers in there and you will get a feel for it. There is no definite answer as to 'how much' error is acceptable.

I'll accept this :)

? Since when did current have any preferences?

In a shortcircuited scenario it has a preference to flow where there are no resistors.

The expression 'No Load' means an Infinite Resistance so the voltage on the Potential Divider would not be affected.

How can "no load" mean infinite resistance? Load by definition means burdens, or something which weighs down on or oppresses! Saying no load means no resistors, not infinite resistance, IMO!

 

[Kholdstare]

A load is useless if current does not flow through it and produce work. In infinite resistance no current flows through - thus "no load".

 

[I like Serena]

Just substitute some numbers in there and you will get a feel for it. There is no definite answer as to 'how much' error is acceptable.

I believe that as a "rule of thumb" 2% error is considered negligible in engineering best practices (sometimes 5%).
However, I can't find such a rule of thumb with google (yet).

 

[jim hardy]

How can "no load" mean infinite resistance? Load by definition means burdens, or something which weighs down on or oppresses! Saying no load means no resistors, not infinite resistance, IMO!

That's wonderful ! I love it ! How logical !

It's electrical jargon - no load is generally accepted to mean unloaded, unburdened, unattached and unencumbered, as opposed to undefined...

 

[sophiecentaur]

How can "no load" mean infinite resistance? Load by definition means burdens, or something which weighs down on or oppresses! Saying no load means no resistors, not infinite resistance, IMO!

This is just history; you just need to adjust your viewpoint to account for someone else's reasoning, I think. No power is delivered to an infinite resistance. The other extreme would, of course, be a Short Circuit, which would be drawn on a diagram as a line joining the two points.

Other possible confusions are "Open Circuit" which means no way through, whereas " Road Open" means traffic can get through. When a switch is 'closed' current can flow but a Road Closed sign means traffic can't. language is a funny thing.

 

[Femme_physics]

I believe that as a "rule of thumb" 2% error is considered negligible in engineering best practices (sometimes 5%).
However, I can't find such a rule of thumb with google (yet).

Thanks, I accept :)

That's wonderful ! I love it ! How logical !

This is just history; you just need to adjust your viewpoint to account for someone else's reasoning, I think. No power is delivered to an infinite resistance. The other extreme would, of course, be a Short Circuit, which would be drawn on a diagram as a line joining the two points.

Hmm... then according to your logic it appears that 'No load" means the load that the CURRENT applies on the Vout. If it's no load, then it means there must be infinite resistance. Does that sound right?

 

[NascentOxygen]

Hmm... then according to your logic it appears that 'No load" means the load that the CURRENT applies on the Vout. If it's no load, then it means there must be infinite resistance. Does that sound right?

Load can be expressed in terms of current, yes. Example, a load of 3 amps.

So "no load" is a load of zero amps. And only if the load is infinite ohms can you have zero current flowing while you supply voltage to it.

 

[Femme_physics]

Load can be expressed in terms of current, yes. Example, a load of 3 amps.

So "no load" is a load of zero amps. And only if the load is infinite ohms can you have zero current flowing while you're supplying voltage to it.

Aha!

Got it!

 

[sophiecentaur]

And as for "how much is an acceptable error?", the answer depends upon the application.
If you wanted to make a reference voltage for measuring some other voltage to within 0.1% (not very common but some people are into that sort of thing) then your 'load' would need to be at least 1000 times the value of the bottom resistor. For use in a domestic room thermostat, you would only be interested in a percent or so - so your precision would reflect this - a higher precision would cost too much in time and effort.

Strictly speaking, the word Load really refers to Power, rather than current. A 2kW load in the US involves twice the current of a 2KW load in the UK.

 

[psparky]

If your truly interested in learning...click on the following link...starting on page 2. I attempt to explain the basics of electrical theory...including your voltage divider. Give it a good read and your understanding will likely go up. Enjoy.

https://www.physicsforums.com/showthread.php?t=588411&page=2

Start with post #30...V=IR

 

[sophiecentaur]

In the same way that you wouldn't expect ot wave your arms about discussing the total on your restaurant bill, simple electronic circuits are far better discussed using Maths. Spreadsheets (even free ones) are so good for evaluating formulae and plotting graphs with varying values of volts, resistance etc. There really is never a good 'chatty' way of describing how all these simple circuits operate. If you don't bite the mathematical bullet with circuits then you may as well accept that you'll never really 'get it'.

 

[psparky]

In the same way that you wouldn't expect ot wave your arms about discussing the total on your restaurant bill, simple electronic circuits are far better discussed using Maths. Spreadsheets (even free ones) are so good for evaluating formulae and plotting graphs with varying values of volts, resistance etc. There really is never a good 'chatty' way of describing how all these simple circuits operate. If you don't bite the mathematical bullet with circuits then you may as well accept that you'll never really 'get it'.

Yes...there is a good way to discuss with chat AND math. I have done that.

Math is great...but even better with a REAL explanation. Teaching while pointing to a formula is lame...dry...and boring. In my opinion.

 

[sophiecentaur]

Unfortunately, we don't have 'real' equivalents in our heads for convolution, cross correlation, Fourier Transforms or, in fact, any of the operations that are vital when dealing with some of even the most elementary processes in Science. We do often 'wave our arms about' from a position of already being familiar with a certain level of the Maths in order to get to the next step, of course. In such a dialogue, however, it may be that only one of the participants is not actually way out of his depth. (Been there)

Imo, it is actually bebasing Physics to think that it can be understood using just a few 'schoolboy' terms and concepts and no Maths. Many of the interchanges I read on these fora are the result of misconceptions which result from the reluctance or inability to appreciate just what the Maths is telling us.

Your 'real' explanation, to my mind, is one that can suit you, personally, but is more of a self-affirming argument and it can be a final step in getting to the next level of (feel -good) understanding. So many people want to be able to leap in with a "So what about this particle nature" or a "The Fourier Series is just . . . . ." and they are very likely just fooling themselves. The proof of the pudding is when they can make the next step of a valid prediction, based on what they feel they know.
I am only too well aware of my limits in this respect.

 

[psparky]

Unfortunately, we don't have 'real' equivalents in our heads for convolution, cross correlation, Fourier Transforms or, in fact, any of the operations that are vital when dealing with some of even the most elementary processes in Science. We do often 'wave our arms about' from a position of already being familiar with a certain level of the Maths in order to get to the next step, of course. In such a dialogue, however, it may be that only one of the participants is not actually way out of his depth. (Been there)

This stuff is great...it's just that it only suits you genuis's. For the rest of us...98% of us...this stuff is totally useless at our jobs in real life. For you 2%'ers...have a blast. You've earned it.

Imo, it is actually bebasing Physics to think that it can be understood using just a few 'schoolboy' terms and concepts and no Maths. Many of the interchanges I read on these fora are the result of misconceptions which result from the reluctance or inability to appreciate just what the Maths is telling us.

I agree to some extent.

Your 'real' explanation, to my mind, is one that can suit you, personally, but is more of a self-affirming argument and it can be a final step in getting to the next level of (feel -good) understanding. So many people want to be able to leap in with a "So what about this particle nature" or a "The Fourier Series is just . . . . ." and they are very likely just fooling themselves. The proof of the pudding is when they can make the next step of a valid prediction, based on what they feel they know.
I am only too well aware of my limits in this respect.

I agree to a certain extend here as well...but me simply laying out the BASICS of EE...is just me trying to lay a foundation for all the rest.

You and I typically disagree on most things...I'm ok with that. You are clearly much older than me and have way more experience. We both make valid points...it's up to the reader to decide who's points' interest them more.

 

[sophiecentaur]

Yebbut it's not the 2% ers. The number of kids who get fair grades at GCSE Maths is up in the 20% and that's all you need to get the idea of putting numbers into formulae and 'believing' that the answer is what you wanted.

There is a fine line between wanting a subject to be treated with respect and just being elitist, I realize that. But some things just aren't accessible without the right tools and it is fooling oneself not to acknowledge that. I keep reading so many "but surely"s from people who are relying on gut feeling about something that geniuses have sweated blood over to get the right answer for. Sometimes it's Andy Wharhol meets Physics - a scientist for 15 minutes. OK and fun for the individual but what sort of message does it give the potentially good Scientist - or the soon to be disappointed one?

Yes - fine with your last comment. All good stuff.;-)

 

[psparky]

Yebbut it's not the 2% ers. The number of kids who get fair grades at GCSE Maths is up in the 20% and that's all you need to get the idea of putting numbers into formulae and 'believing' that the answer is what you wanted.

There is a fine line between wanting a subject to be treated with respect and just being elitist, I realize that. But some things just aren't accessible without the right tools and it is fooling oneself not to acknowledge that. I keep reading so many "but surely"s from people who are relying on gut feeling about something that geniuses have sweated blood over to get the right answer for. Sometimes it's Andy Wharhol meets Physics - a scientist for 15 minutes. OK and fun for the individual but what sort of message does it give the potentially good Scientist - or the soon to be disappointed one?

Yes - fine with your last comment. All good stuff.;-)

Fair enough...good. I too like a mathematical proof...sometimes I do them on my own. All good. I also like a "laymen's" explanation as well...

Most of my statements are taken from tutoring engineers for the F.E and P.E. tests. (Two tests you need to be a "Professional Engineer" in the USA.) I see where there weaknesses are...and they start the weaknesses at rock bottom! That's why I wrote up those basics...now I can just hand them a hand out with a brief explanation!

When studying for those tests...most people have very limited time available. So you need to pick your battles on what you study for.

A few topics I tell them to throw out the window for the tests...convolution, Fourier transforms and analysis, signal processing...and a few others as well. In my opinion...those are the tougher subjects...they just don't have the time to master them. I have never mastered them...simply because I knew I could by without them and I know my work will never use them...and their subject manner is a super small part of the tests...or even college...(at least in USA)

Now perhaps I just never had the right teacher for them. If someone just throws a bunch of math in front of me...I'll never even look at it. But if someone where to start with a laymen's explanation of each...perhaps I might get interested. Just sayin...who wants to teach these subjects? Go on...if you teach it...I'll listen. Perhaps I am missing out on some of the greatest things ever in the history of the world!

 

[sophiecentaur]

I wouldn't say that you are a typical 'layman' so you are in a better position to have opinions about these things than yer average Jo public. You make some very good points which are really worth reading.

 

[psparky]

I wouldn't say that you are a typical 'layman' so you are in a better position to have opinions about these things than yer average Jo public. You make some very good points which are really worth reading.

Thank you. Your kind words are most appreciated.

 

[sophiecentaur]

Feel free to chuck more brickbats and ya-boos my boy!

 

[psparky]

Will do.

So who is going to teach me about convolution? In a way that I will understand it backwards and forwards...not to mention anyone else who may be reading your amazing lecture worthy of a Pulitzer prize...lol

Maybe start with "what it is"...and "why we need it". Then go into math.

Anyone up for the challenge?

 

[sophiecentaur]

The one word that sticks with me about Convolution is 'Overlap'. If you sweep one function across another function and see how much they overlap during that sweep then that is one function convolved with the other. When two single-peaked functions are convolved with each other, the result gives a 'sharper peak' in the position where the two separate peaks coincide.
When dealing with periodic signals, you get an interesting relationship between the frequency and the time domain. Convolution in the time domain becomes (by the Fourier transform) multiplication in the frequency domain. (there are a lot of little nuggets like that)
Actually, there is an alternative to me blathering on. Try this link. He talks around the processes as well as giving the Maths.

It may not help and it was all years ago for me but I remember when it was pointed out that
∫Cos(ω1t) Cos(ω2t)dt (-∞ to ∞) is zero when ω1≠ω2

This is the basis of understanding the way Fourier works in giving you the 'amount' of a particular frequency component in any general signal - because that above operation (the core of Fourier) only gives a non zero value for the frequency of interest. When I saw it in this way, the Fourier business made sense.

 

[psparky]

It may not help and it was all years ago for me but I remember when it was pointed out that
∫Cos(ω1t) Cos(ω2t)dt (-∞ to ∞) is zero when ω1≠ω2

This is the basis of understanding the way Fourier works in giving you the 'amount' of a particular frequency component in any general signal - because that above operation (the core of Fourier) only gives a non zero value for the frequency of interest. When I saw it in this way, the Fourier business made sense.

So the integral (area under curve) of first frequency multiplied by the deravitive (rate of change) of the second frequency under any time period (same time period for both)...will yield zero under all cases except when the two frequencies are equal.

Is that true? I assume you are talking about identical wave forms with different frequencies?

If so...pretty cool...now what does that mean or do for us?

 

[sophiecentaur]

So the integral (area under curve) of first frequency multiplied by the deravitive (rate of change) of the second frequency under any time period (same time period for both)...will yield zero under all cases except when the two frequencies are equal.

Is that true? I assume you are talking about identical wave forms with different frequencies?

If so...pretty cool...now what does that mean or do for us?

Is that your interpretation of the expression I wrote?. What I wrote is the integral (wrt t) of a product over all time of the two Cos functions. The two waveforms are sinusoidal (Cos waves) and their frequencies are w1 and w2. This is very standard Maths notation. If it is a problem for you then I think you may be right about never getting convolution.

 

[DragonPetter]

I think convolution is easier to understand if you consider it in the discrete(digital) version, and do some real examples.